ISC Class 12 Physics Chapter Wise Questions with Solutions

Explanation: This question focuses on the applications and limitations of Huygens’ wave theory of Light. Huygens proposed that every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is formed by the envelope of these wavelets. Using this geometrical idea, several optical phenomena can be understood successfully.

The important concepts connected with this topic include reflection, refraction, and Diffraction of Light. Reflection can be explained because the secondary wavelets reproduce the law of reflection geometrically. Refraction is also derived through the change in wave velocity in different media. Diffraction becomes understandable because wavelets spread into regions beyond obstacles and apertures.

However, some optical effects require deeper analysis involving wavelength separation and interaction of Light with Matter. Such effects are beyond the simple geometrical construction of wavefronts. Huygens’ principle mainly provides a method for tracing propagation of waves rather than explaining every optical phenomenon completely.

A useful comparison is water waves passing through a small opening. The spreading of waves after the opening resembles Diffraction and is easily visualized through secondary wavelets.

Thus, the question tests understanding of which phenomena are directly explained by wavefront construction and which require additional physical theories beyond Huygens’ original model.

Explanation: This question examines the relationship between wavefronts and the direction in which Light propagates. In wave Optics, a wavefront represents a surface on which all particles vibrate in the same phase at a particular instant. The direction of energy transfer is always perpendicular to this surface.

The key idea here is the geometrical representation of Light propagation. When perpendicular lines are drawn from a wavefront, they indicate the direction in which the disturbance travels. These lines help in understanding reflection, refraction, and image formation in Optics.

To analyze the concept, imagine a stone dropped in still water. Circular ripples spread outward, and if arrows are drawn perpendicular to the ripples, they show the direction of wave travel. The same principle applies to Light waves. In a plane wavefront, these perpendicular lines remain parallel, while in a spherical wavefront they either converge or diverge depending on the source.

This relationship between wavefronts and perpendicular propagation paths is fundamental in wave Optics and helps simplify the study of optical systems. The question mainly checks conceptual clarity about how wavefront geometry describes the motion and transmission of Light energy through space.

Explanation: This question checks the understanding of how angles are measured during the reflection of Light from a surface. In geometrical Optics, reflection occurs when a Light wave strikes a surface and returns into the same medium. To describe this process correctly, specific reference lines are used.

The important concept involved is the normal line, which is an imaginary line drawn perpendicular to the reflecting surface at the point where light strikes. Both the incident angle and reflected angle are always measured with respect to this normal, not with the surface itself. This convention helps maintain consistency in the law of reflection.

To reason through the problem, imagine a smooth mirror placed vertically. A light ray strikes the mirror and bounces back. If the angle were measured from the surface directly, values would vary and become confusing. Measuring from the normal provides a standard method and makes the incident and reflected angles comparable. This is why the law of reflection states that both these angles are equal.

A similar idea appears in billiards, where the path of a ball after striking the side cushion depends on its direction relative to a perpendicular reference line.

Thus, the question tests understanding of the geometrical definition used in reflection and the importance of the normal line in optical measurements.

Explanation: This question focuses on the properties of light during reflection from a mirror. Reflection is a phenomenon in which light returns into the same medium after striking a surface. While the direction of propagation changes, certain wave properties remain unaffected because the medium itself does not change.

Important properties of light include wavelength, frequency, speed, and phase. Since reflected light continues to travel in the same medium, its speed remains unchanged. Frequency also remains constant because it depends on the source producing the light. As wavelength depends on both speed and frequency, it also remains unchanged in the same medium.

However, during reflection, especially from denser surfaces, a change related to wave orientation can occur. This effect is associated with the wave’s phase. The reflected wave may experience a shift depending on the nature of the reflecting surface and boundary conditions.

Consider a stretched rope tied at one end. When a pulse reflects from the fixed end, it returns inverted. This inversion resembles a phase-related change in wave behavior during reflection.

The question therefore examines which physical quantity may undergo modification during reflection while other fundamental properties remain constant in the same medium.

Explanation: This question is related to refraction and the behavior of wavefronts when light passes from one medium to another. Refraction occurs because the speed of light changes when it enters a different medium. The geometry of wavefronts helps explain this bending of light using Huygens’ principle.

The key concepts involved are incident wavefront, refracted wavefront, and the relation between angles and wave velocities. Normally, when light changes medium, the refracted wavefront changes its orientation due to a difference in speed. However, if the angle of incidence and refraction are equal, the wavefront geometry suggests that there is no bending at the interface.

To understand this, imagine a row of marching soldiers entering another region. If every soldier continues forward without turning, the front line maintains the same orientation and shape. Similarly, when the incident and refracted angles are identical, the sizes of corresponding wavefront portions remain unchanged.

This situation usually indicates equal propagation conditions in both media. Since the wavefront does not tilt differently, its dimensions relative to the incident wavefront remain unchanged as well.

Thus, the question examines the connection between equal angles in refraction and the geometrical relationship between incident and refracted wavefronts.

Explanation: This question deals with the geometrical nature of plane wavefronts in wave Optics. A wavefront is the surface joining all points vibrating in the same phase. Depending on the source and distance, wavefronts may be spherical, cylindrical, or plane.

The radius of curvature describes how strongly a wavefront bends. Spherical wavefronts have finite radii because they originate from nearby point sources. As the distance from the source increases, the curvature gradually decreases, and the wavefront becomes flatter. A perfectly plane wavefront is considered to have no curvature.

To analyze this, think of a very large circle. If only a tiny portion of its edge is observed, it appears almost straight. In Optics, light coming from extremely distant sources behaves similarly. The wavefront becomes so large that its curvature is practically undetectable, giving rise to a plane wavefront.

Because a plane wavefront is flat, its radius of curvature is treated mathematically in a special way. Instead of having a finite value, it is considered extremely large.

This question therefore checks understanding of the geometrical interpretation of plane wavefronts and how curvature behaves during wave propagation.

Explanation: This question examines the fundamental property of waves and what exactly gets transferred during wave motion. Waves are disturbances that travel through a medium or space while carrying certain physical effects from one location to another.

The important idea here is that particles of the medium generally do not travel along with the wave over long distances. Instead, they oscillate about their mean positions while the disturbance continues moving forward. This principle applies to sound waves, water waves, and electromagnetic waves.

To understand this clearly, consider a ripple moving across a pond. The water itself does not travel from one side of the pond to the other. Individual water particles simply move up and down while the disturbance spreads outward. Yet something useful is transferred across the pond, showing the wave carries a physical quantity through motion.

The same principle is observed in a stretched rope. When one end is shaken, the disturbance reaches the other end even though the rope particles themselves remain approximately in place.

Thus, the question tests understanding of the basic purpose of wave motion and distinguishes between Transport of Matter and Transport associated with the propagation of a disturbance.

Explanation: This question is based on Huygens’ principle and the role of secondary wavelets in wave propagation. According to Huygens, every point on a wavefront behaves as a source of tiny secondary disturbances. The new wavefront at a later instant is obtained by drawing the envelope touching all these wavelets.

The central concept here is geometrical construction of wavefronts. Huygens’ idea does not directly calculate numerical quantities like wavelength or speed but provides a graphical method to determine how wavefronts move and change shape during propagation.

To reason through the concept, imagine many small stones dropped simultaneously along a line in water. Each stone creates circular ripples, and the outer boundary formed by all ripples together represents the new advancing wavefront. This helps explain reflection and refraction geometrically.

The principle is extremely useful in predicting the future position of waves after interacting with surfaces or changing media. It gives a visual understanding of wave propagation without requiring particle-based assumptions.

Therefore, this question checks whether the learner understands the actual purpose of secondary wavelets and how they help construct new wavefronts in wave Optics.

Explanation: This question tests the basic statement of Huygens’ principle in wave Optics. The principle was introduced to explain how waves propagate through space and how new wavefronts are continuously formed during motion.

The key idea is that every point on a given wavefront acts like a fresh source of disturbance. These tiny disturbances spread outward in all directions with the same speed as the original wave. The surface tangent to all these secondary wavelets at a later time gives the next wavefront.

To understand this process, imagine spectators in a stadium performing a wave. Each person, after receiving the disturbance, creates a new disturbance by standing up and sitting down. The overall moving pattern resembles the formation of successive wavefronts from individual points.

This principle successfully explains many optical phenomena such as reflection and refraction through geometrical reasoning. It also establishes the relationship between wavefronts and the direction of wave propagation.

The question mainly checks conceptual clarity about the fundamental assumption made in Huygens’ theory and the role of individual points on a wavefront during the transmission of light waves.

Explanation: This question concerns the nature of wavefronts produced by very distant light sources. The Sun emits light in all directions, producing spherical wavefronts near the source. However, the appearance of these wavefronts changes significantly when observed at extremely large distances.

The important concept here is curvature. A spherical wavefront has a finite radius near the source, but as distance increases, its curvature decreases. By the time sunlight reaches Earth, the radius of the spherical wavefront becomes enormously large compared to Earth’s dimensions.

To visualize this, think of standing on a huge football field and observing a tiny curved section of an enormous circle. That small section appears almost perfectly straight. Similarly, the small portion of the Sun’s wavefront reaching Earth behaves like a flat surface.

This approximation greatly simplifies optical calculations and is commonly used in wave Optics and geometrical Optics. It helps explain why sunlight rays arriving at Earth are considered nearly parallel.

Thus, the question checks understanding of how distance affects the shape and curvature of wavefronts emitted by distant celestial sources.

Explanation: This question deals with the formation of different types of wavefronts depending on the shape of the light source. In Wave Optics, the geometry of the source determines whether the wavefront becomes spherical, cylindrical, or plane.

A cylindrical wavefront is formed when light originates from a linear source rather than a point source. Every point along the line source emits secondary wavelets, and together they produce a wavefront shaped like a cylinder expanding outward.

To understand this better, imagine a long glowing tube placed vertically or horizontally. Light spreads outward around the length of the tube, forming curved surfaces around the axis of the source. Unlike spherical wavefronts, which spread in all directions equally, cylindrical wavefronts spread mainly around the line source.

The orientation of the line source may vary, but the nature of the resulting wavefront remains cylindrical as long as the source behaves like a straight luminous line.

This question therefore checks understanding of the connection between source geometry and the shape of the produced wavefront in Wave Optics.

Explanation: This question is related to Diffraction and the behavior of light after passing through a narrow opening. According to Huygens’ principle, every point on the slit acts as a source of secondary wavelets that spread outward into the surrounding space.

The shape of the resulting wavefront depends on the dimensions of the slit and the arrangement of the source. A very narrow slit effectively behaves like a small line source, causing the emerging wavefront to spread outward in a curved manner.

To understand this physically, imagine ocean waves approaching a narrow gap between rocks. After passing through the gap, the waves spread into the region beyond instead of continuing only in a straight line. Light behaves similarly due to its wave nature.

The outgoing wavefront from a narrow slit therefore differs from the original incoming wavefront. This spreading phenomenon forms the basis of Diffraction and demonstrates that light does not always travel as perfectly straight rays.

Thus, the question tests understanding of how wavefront geometry changes when light passes through a narrow aperture and how Huygens’ secondary wavelets explain this behavior.

Explanation: This question examines the geometrical relationship between wavefronts and the direction of propagation of light. In Wave Optics, a wavefront is an imaginary surface joining all points that vibrate in the same phase at a particular instant.

The important concept here is that energy always travels perpendicular to the wavefront. Therefore, if a line is drawn normal to the wavefront in the direction of motion, it represents the path along which light travels. This idea is essential in understanding reflection, refraction, and optical image formation.

To visualize this, imagine circular ripples produced on the surface of water. The ripples themselves represent wavefronts, while arrows drawn outward perpendicular to the ripples indicate the direction in which the disturbance moves. In optics, the same principle applies to light waves.

For plane wavefronts, these perpendicular lines remain parallel to one another. For spherical wavefronts, they either diverge outward or converge inward depending on the source configuration.

Thus, the question tests the learner’s understanding of how wavefront geometry is connected with the actual propagation path of light in Wave Optics.

Explanation: This question focuses on the geometrical relation between wavefronts and light rays. A wavefront is a surface connecting points vibrating in the same phase, whereas a ray represents the direction in which light energy propagates.

The key concept involved is the orientation between these two entities. Since light travels normal to the wavefront, rays are always drawn perpendicular to the wavefront surface. This relationship forms the basis of geometrical optics and helps explain optical laws mathematically.

To understand this clearly, consider ripples spreading on the surface of water. The circular ripple itself acts like a wavefront. If arrows are drawn outward from the center crossing the ripple at right angles, those arrows represent the direction of propagation. The same idea applies to light waves.

In spherical wavefronts, rays either diverge from or converge toward a point. In plane wavefronts, all rays remain parallel because the wavefront is flat.

This question therefore checks conceptual clarity regarding how rays and wavefronts are related geometrically and why their orientation is important in the study of optical phenomena.

Explanation: This question deals with the behavior of wave normals for different types of wavefronts. A wave normal is a line drawn perpendicular to the wavefront in the direction of propagation of the wave. The pattern of these normals depends entirely on the shape of the wavefront.

For spherical wavefronts produced by a point source, the wavefronts expand outward in all directions. Therefore, the wave normals spread outward from the center. This creates a diverging pattern because every ray moves away from the source point.

In contrast, a plane wavefront represents a flat surface usually produced when the source is extremely far away. Since the surface is flat, all perpendiculars drawn on it remain equally spaced and never meet. Hence, the wave normals remain parallel.

An easy analogy is sunlight reaching Earth. Because the Sun is extremely far away, the arriving rays appear parallel. On the other hand, light from a nearby bulb spreads outward in different directions.

Thus, the question checks understanding of how the geometry of the wavefront determines the arrangement and behavior of wave normals in Wave Optics.

Explanation: This question is based on the shape of wavefronts produced by different light sources. A point source emits light equally in all directions, causing the disturbance to spread outward symmetrically through space.

The important concept here is that when the source is at a finite distance, the emitted wavefronts possess noticeable curvature. Every successive wavefront forms a curved surface centered around the source point. This geometry results from equal spreading in all directions.

To visualize this, imagine inflating a balloon from a single point. As the balloon expands, every layer forms a curved surface surrounding the center. Light from a nearby point source behaves similarly, producing expanding curved wavefronts.

The curvature becomes smaller only when the source is extremely distant. In that case, a tiny observed portion appears almost flat. But at finite distances, the curvature remains significant and cannot be ignored.

This question therefore examines the relationship between source distance and the resulting shape of the wavefront produced in Wave Optics.

Explanation: This question examines how the curvature of a spherical wavefront changes when the source is extremely far away. Light emitted from a point source normally forms spherical wavefronts, but their appearance depends strongly on the observer’s distance from the source.

The key concept is that the radius of curvature increases continuously with distance. When the source is effectively at an infinite distance, the radius becomes enormously large. As a result, a very small portion of that curved surface appears nearly flat.

A useful analogy is Earth itself. Although Earth is spherical, a small area viewed locally appears flat because the curvature is too small to notice over short distances. Similarly, a tiny section of a huge spherical wavefront behaves like a flat surface.

This approximation is extremely important in optics because it simplifies mathematical treatment of distant sources such as sunlight or starlight. The rays associated with such wavefronts are treated as parallel.

Thus, the question tests understanding of how very large spherical wavefronts can locally behave like flat wavefronts due to extremely small curvature.

Explanation: This question involves the behavior of light passing through a convex lens when the source is placed at a special position. A convex lens refracts light rays in such a way that their direction changes depending on the source location relative to the focal point.

The key principle here is that light emerging from the focus of a convex lens undergoes refraction that makes the outgoing rays parallel to each other. Since wavefronts are always perpendicular to the direction of rays, the shape of the resulting wavefront changes accordingly.

To visualize this, imagine water flowing outward from a point and then being redirected by a carefully shaped channel so that all streams become parallel. Similarly, the lens converts the diverging rays from the point source into parallel rays.

Parallel rays correspond to a flat advancing wavefront rather than a curved one. This property is widely used in optical instruments such as searchlights and telescopes where parallel beams are required.

Thus, the question checks understanding of the relation between focal points, ray behavior, and the type of wavefront formed after refraction through a convex lens.

Explanation: This question examines the response of a direct current measuring instrument when Alternating Current flows through the circuit. A D.C. ammeter is designed to measure current flowing steadily in one direction, whereas Alternating Current continuously changes direction with time.

The important concept here is the average effect of Alternating Current over a complete cycle. In an A.C. circuit operating at 50 Hz, the current reverses direction fifty times each second. During one half-cycle, the current flows in one direction, and during the next half-cycle, it flows equally in the opposite direction.

Because of this rapid reversal, the magnetic effects inside the D.C. ammeter oppose each other. Over a complete cycle, the positive and negative contributions balance out. As a result, the pointer does not experience a steady deflecting force.

A similar situation occurs if a person walks equal distances forward and backward repeatedly. Even though movement occurs continuously, the average displacement after each full cycle becomes zero.

Thus, the question tests understanding of why ordinary D.C. measuring instruments are unsuitable for detecting Alternating Current directly.

Explanation: This question concerns the design principle of an ammeter and its effect on an electric circuit. An ammeter is connected in series with the circuit to measure the current flowing through it. Since series components affect total resistance, the instrument must be carefully designed.

The key concept is that adding extra resistance in series changes the current of the entire circuit. If the ammeter had high resistance, it would oppose current flow significantly and alter the very quantity it is supposed to measure. Therefore, the instrument should offer minimal opposition.

To understand this practically, imagine water flowing through a pipe. If a measuring device inserted into the pipe creates a narrow obstruction, the water flow itself changes. A proper measuring instrument should disturb the system as little as possible.

A low-resistance ammeter allows almost the same current to pass as would flow without the instrument. This ensures accurate measurement while preventing unnecessary energy loss and heating.

Thus, the question checks conceptual understanding of why ammeters are designed with extremely small resistance values in electrical measurements.

Explanation: This question is based on extending the range of an ammeter using a shunt resistance. An ammeter is usually constructed from a galvanometer, which can safely carry only a limited amount of current. To measure larger currents, an additional low resistance is connected in parallel.

The important concept involved is current division in parallel circuits. Since parallel branches have the same potential difference, most of the current is diverted through the low-resistance shunt while only a safe fraction passes through the measuring instrument.

To analyze the situation, suppose the original meter safely measures current up to I ampere. If the range is increased to nI amperes, the extra current beyond the original limit must bypass the instrument through the shunt branch. Using Ohm’s law and parallel current relations, the required resistance can be expressed in terms of G and n.

This method is widely used in practical electrical instruments because it protects sensitive measuring coils from excessive current while enabling measurement of large values.

Thus, the question tests understanding of shunt resistance and the mathematical principle used for extending the range of an ammeter.

Explanation: This question involves converting a galvanometer into an ammeter using a shunt resistance. A galvanometer is a sensitive instrument that can detect only small currents. To measure larger currents safely, a low resistance is connected in parallel so that most current bypasses the galvanometer.

The important concept here is current division in a parallel circuit. Since both the galvanometer and shunt have the same potential difference across them, the currents through the two branches are inversely proportional to their resistances.

To solve such problems, first determine the voltage across the galvanometer at full-scale deflection using Ohm’s law. The same voltage exists across the shunt resistance. From this, the current through the shunt branch can be calculated. The total measurable current becomes the sum of galvanometer current and shunt current.

A practical analogy is a traffic diversion system where only a small number of vehicles pass through a narrow inspection lane while the majority are redirected through a wider bypass route.

Thus, the question tests understanding of shunt resistance and the principle of extending current-measuring capacity in electrical instruments.

Explanation: This question is based on extending the current range of a sensitive measuring instrument using a shunt resistance. A meter giving full-scale deflection at a very small voltage can carry only a limited amount of current safely. To measure larger currents, most of the current must bypass the instrument.

The important concept here is the use of a low-resistance shunt connected in parallel with the meter. Since both parallel branches have the same potential difference across them, the current divides according to their resistances. The meter branch carries only its safe current while the remaining current passes through the shunt.

To analyze the problem, first determine the full-scale current of the meter using Ohm’s law from the given resistance and voltage values. Then compare this safe current with the desired total current range. The difference must pass through the shunt branch. Using parallel circuit relations, the required shunt resistance can be calculated.

A useful analogy is a dam with a small controlled channel and a large bypass path carrying excess water flow safely.

Thus, the question checks understanding of current division and practical range extension of ammeters using shunt resistances.

Explanation: This question examines how a galvanometer is converted into an ammeter by using a shunt resistance. A galvanometer is highly sensitive and can carry only a small fraction of the total current. Therefore, a parallel low-resistance path is used so that most of the current bypasses the galvanometer safely.

The important principle involved is current division in parallel branches. Since both the galvanometer and shunt experience the same potential difference, the current through each branch becomes inversely proportional to its resistance.

To reason through the problem, note that only a small percentage of the total current passes through the galvanometer while the remaining larger portion flows through the shunt. Using the ratio of currents and the equality of voltage across parallel branches, the relation between galvanometer resistance and shunt resistance can be derived mathematically.

A simple comparison is a highway toll system where only a few vehicles pass through a narrow lane while the majority use a broader diversion route to avoid congestion.

Thus, the question tests understanding of parallel current distribution and the design principle behind extending the range of current-measuring instruments.

Explanation: This question concerns the use of a shunt resistance with a galvanometer to measure larger currents. A moving coil galvanometer is a delicate instrument and cannot safely carry the entire circuit current. Therefore, a low-resistance shunt is connected in parallel to divert excess current.

The key concept here is current division in a parallel circuit. Because the galvanometer and shunt are connected across the same two points, they experience equal potential difference. The branch carrying larger current must therefore possess smaller resistance.

To analyze the problem, first identify the fraction of total current intended to pass through the galvanometer. The remaining current flows through the shunt branch. By applying the condition of equal voltage across parallel branches and using Ohm’s law, the required shunt resistance can be obtained from the ratio of currents.

An analogy can be made with traffic management. If only a small fraction of vehicles are allowed through a narrow inspection lane, the remaining majority must move through a wider bypass route with lower resistance to flow.

Thus, the question checks understanding of current division and practical modification of galvanometers for high-current measurements.

Explanation: This question evaluates conceptual understanding of ammeter construction and the role of shunt resistance in extending measuring range. An ammeter is designed to measure current while introducing minimal disturbance into the circuit.

The important concept involved is that increasing the range of an ammeter requires diverting a larger fraction of current away from the sensitive galvanometer coil. This is achieved by connecting an additional low-resistance shunt in parallel with the instrument.

To analyze the statements, recall that parallel combination generally decreases equivalent resistance. When a smaller shunt is connected, more current bypasses the galvanometer, allowing measurement of larger currents safely. Therefore, the behavior of total ammeter resistance must be examined carefully instead of assuming it increases automatically with range.

A practical analogy is widening a traffic diversion route. As more traffic is redirected through wider side paths, the overall resistance to flow becomes smaller rather than larger.

This question mainly tests logical interpretation of electrical instrument design and whether the explanation provided correctly supports the related statement about range extension.

Explanation: This question is related to the effect of inserting an ammeter into an electrical circuit. Since an ammeter is used to measure current, it must be connected in series so that the same current flows through both the instrument and the circuit components.

The key principle here is that any component connected in series contributes additional resistance to the total circuit resistance. However, a properly designed ammeter possesses extremely low resistance so that its effect on the circuit remains negligible.

To understand this, imagine water flowing through a pipe. If a measuring device with even slight obstruction is inserted, the total opposition to water flow increases slightly. Similarly, the circuit resistance technically increases when the ammeter is connected, although the increase is very small due to its low resistance.

This design ensures that the current in the circuit remains almost unchanged during measurement. Otherwise, the instrument would alter the quantity it is supposed to measure.

Thus, the question checks conceptual understanding of series circuits, total resistance, and why ammeters are designed with very low resistance values.

Explanation: This question focuses on the ideal characteristics required for an ammeter. An ammeter measures the current flowing through a circuit and is always connected in series so that the same current passes through the instrument.

The important concept is that any resistance inserted in series affects the circuit current. If the ammeter had noticeable resistance, it would reduce the current and produce inaccurate measurements. Therefore, an ideal measuring instrument should disturb the original circuit conditions as little as possible.

To reason through this, imagine measuring the flow of water in a pipe. If the measuring device blocks the pipe significantly, the water flow itself changes. A perfect measuring device would allow water to pass without any obstruction. The same principle applies to electrical current.

In practical instruments, achieving absolutely no resistance is impossible, but the resistance is kept extremely small compared to the circuit resistance. This minimizes energy loss and heating while ensuring accurate readings.

Thus, the question tests understanding of why the theoretical resistance of an ideal ammeter is taken to be the minimum possible value in electrical measurements.

Explanation: This question examines how a galvanometer is modified to measure larger currents. A galvanometer by itself is a highly sensitive device capable of detecting only small currents. To make it suitable for practical current measurements, a low resistance called a shunt is connected in parallel.

The key concept here is current division. In a parallel arrangement, only a small fraction of the total current passes through the galvanometer while the remaining larger portion flows through the shunt branch. This protects the sensitive coil from damage.

To analyze the arrangement, remember that instruments are classified according to the quantity they measure. A galvanometer detects small currents, but after adding the shunt, the combined device becomes capable of measuring stronger currents directly in a circuit.

A useful analogy is a narrow bridge supported by a larger bypass road. Most vehicles use the wider route, allowing the delicate bridge to remain safe while still participating in the traffic system.

Thus, the question checks understanding of how practical current-measuring instruments are constructed from sensitive galvanometers using parallel shunt resistances.

Explanation: This question concerns the method used to extend the measuring capacity of an ammeter. Since the internal galvanometer of an ammeter can carry only limited current safely, modifications are required when larger currents need to be measured.

The important concept involved is the use of a shunt resistance connected in parallel with the instrument. A shunt provides an alternate low-resistance path through which most of the current can flow. As a result, only a safe fraction passes through the measuring coil.

To reason through the situation, consider what happens when the shunt resistance becomes smaller. A lower-resistance branch attracts more current according to current division rules. Therefore, the instrument can now handle a larger total current without damaging the galvanometer.

A practical analogy is widening a diversion road during heavy traffic. The broader the side path, the more vehicles it can carry away from the main narrow lane.

Thus, the question tests understanding of how shunt resistance controls current distribution and enables extension of an ammeter’s operating range.

Explanation: This question is based on increasing the current range of an ammeter using a parallel shunt resistance. The given ammeter can safely measure only up to a certain current, so additional current must bypass the instrument through another low-resistance path.

The important principle involved is that in a parallel circuit, both branches have the same potential difference across them. Since the shunt resistance is very small, most of the total current flows through it while only the rated current passes through the original ammeter.

To analyze the problem, first determine the voltage drop across the ammeter at its maximum safe current using Ohm’s law. Then calculate how much extra current must pass through the shunt branch to extend the range to the desired value. Using the equality of potential difference across parallel branches, the required shunt resistance can be determined.

An analogy can be made with a small bridge supported by a wide alternate route that carries most heavy vehicles during traffic overload.

Thus, the question checks understanding of shunt-based range extension and current division in electrical measuring instruments.

Explanation: This question combines the principles of voltmeters, ammeters, and conversion of measuring instruments. A voltmeter generally has high resistance and is designed to measure potential difference. To use it differently for current measurement, suitable circuit modifications are required.

The important concept here is that converting a voltage-measuring device into a current-measuring instrument involves providing a low-resistance path so that large current can flow without damaging the meter. This is achieved using a parallel arrangement.

To reason through the problem, first determine the maximum current the voltmeter can safely carry from its rated voltage and resistance using Ohm’s law. Since the desired current range is much larger, most of the current must bypass the meter through another resistance connected appropriately. Using current division principles, the required resistance value and its connection type can be identified.

A practical analogy is diverting most water flow through a broad channel while allowing only a limited controlled amount through a measuring pipe.

Thus, the question checks understanding of instrument conversion, Ohm’s law, and the use of shunt resistances in electrical measurement systems.

Explanation: This question is based on the construction of an ammeter using a galvanometer and a shunt resistance. A galvanometer is a sensitive instrument that can detect only small currents. To convert it into an ammeter, a low resistance is connected in parallel so that most of the current bypasses the galvanometer.

The important concept here is equivalent resistance in parallel combinations. Since the galvanometer resistance is comparatively large and the shunt resistance is small, the combined resistance of the arrangement becomes much lower than either branch individually.

To analyze the situation, use the formula for equivalent resistance of two resistors connected in parallel. Because current prefers the path of lower resistance, the effective resistance of the complete ammeter becomes dominated mainly by the shunt branch.

A practical analogy is a narrow road connected alongside a broad highway. Most traffic naturally moves through the easier route, reducing the overall resistance to movement.

Thus, the question checks understanding of parallel resistance combinations and why practical ammeters possess very small effective resistance values.

Explanation: This question involves combined application of Ohm’s law, internal resistance of a cell, and the effect of measuring instruments in a circuit. The ammeter is connected in series while the voltmeter is connected across the resistor to measure potential difference.

The important concept here is that practical voltmeters have finite resistance and therefore draw some current from the circuit. Because of this additional current branch, the total current measured by the ammeter differs from the current through the resistor alone.

To analyze the problem, first determine the total resistance of the circuit using the emf and measured current. Then subtract the known series resistances such as the resistor, internal resistance, and ammeter resistance. The remaining effect is due to the parallel combination involving the voltmeter. Applying parallel resistance relations allows determination of the voltmeter resistance.

A similar situation occurs in water flow systems where a measuring branch diverts a small amount of Fluid from the main pipeline.

Thus, the question tests understanding of practical electrical instruments and how their non-ideal resistances influence circuit behavior.

Explanation: This question examines how stretching a wire affects its electrical resistance. Resistance depends on the material, length, and cross-sectional area of the conductor. When a wire is stretched, both its dimensions change even though the total volume remains constant.

The key concept involved is the relation R=ρ
A
L
​

, where resistance is directly proportional to length and inversely proportional to area. Since density remains unchanged, the volume of the wire remains constant during stretching.

To analyze the situation, note that increasing the length several times causes the cross-sectional area to decrease proportionally to maintain constant volume. Because resistance depends on both increased length and decreased area, the overall resistance rises significantly. By comparing the new resistance expression with the original one, the change in resistance relative to the initial resistance can be obtained.

A practical analogy is stretching a rubber tube carrying water. As the tube becomes longer and thinner, the flow faces greater opposition.

Thus, the question tests understanding of dimensional changes and their combined effect on electrical resistance.

Explanation: This question is related to Kirchhoff’s loop law, which is widely used in analyzing electrical circuits. Kirchhoff’s second law states that the algebraic sum of all potential differences and electromotive forces around a closed loop is zero.

The important principle behind this law is the conservation of a fundamental physical quantity. As charge moves around a complete closed circuit, energy supplied by sources such as batteries must equal the energy lost across resistors and other components.

To understand this, imagine a cyclist riding around a circular track with hills and slopes. Any energy gained while descending must balance the energy required to climb back to the original level. Similarly, in an electric loop, total energy gain and loss balance each other exactly.

This law is extremely useful for solving complex circuits containing multiple loops and sources. It ensures that electrical calculations remain consistent with basic physical principles governing nature.

Thus, the question checks understanding of the physical foundation underlying Kirchhoff’s second law and its connection with conservation principles in circuit analysis.

Explanation: This question is based on Kirchhoff’s loop law, which describes the behavior of electrical quantities in a closed circuit. According to this law, all voltage gains and voltage drops around a complete loop must balance one another.

The key concept involved is the relation between electromotive force and potential drops across resistors. When current flows through a resistor, energy is lost in the form of Heat, producing a voltage drop equal to the product of current and resistance. Sources like batteries provide energy to compensate for these losses.

To analyze the situation, imagine traveling around a circular route with both upward and downward slopes. If you return to the starting point, the total change in height becomes zero because all increases and decreases cancel each other. Electrical loops behave similarly in terms of energy changes.

Mathematically, the algebraic addition of all emf values and voltage drops in a complete loop must satisfy a balance condition. This ensures consistency with the conservation principles governing electrical energy.

Thus, the question tests conceptual understanding of Kirchhoff’s loop law and how voltage gains and losses behave in closed electrical circuits.

Explanation: This question concerns Kirchhoff’s junction law, which is applied at points in a circuit where several branches meet. A junction is a location where electric current splits into different paths or combines from multiple branches.

The important principle behind this law is that electric charge cannot accumulate continuously at a junction in a steady circuit. Therefore, the total current entering the junction must equal the total current leaving it.

To understand this, imagine water flowing through interconnected pipes meeting at a single point. If more water entered than left, water would begin collecting at the junction. In a steady system, such accumulation does not occur, so inflow and outflow remain balanced.

This idea directly reflects a fundamental conservation principle in Physics. Kirchhoff’s junction law is extremely important in solving complex electrical networks because it relates currents in different branches systematically.

Thus, the question checks understanding of the physical basis of current conservation at circuit junctions and the law that governs current distribution in electrical networks.

Explanation: This question involves parallel resistance combinations and the effect of removing one branch from the circuit. When two resistors are connected in parallel, the effective resistance becomes smaller than either individual resistance because current gets multiple paths to flow.

The important concept here is the formula for equivalent resistance in parallel arrangements. Initially, both wires contribute to reducing the total resistance. After one wire breaks, only the remaining resistor determines the effective resistance of the circuit.

To analyze the problem, first identify the resistance of the unbroken wire from the new effective resistance value after the break occurs. Then substitute this known value into the parallel resistance equation corresponding to the original arrangement. Solving the equation gives the resistance of the broken wire.

A useful analogy is two parallel roads connecting the same destinations. If one road closes, all traffic must pass through the remaining road, increasing the effective difficulty of travel.

Thus, the question tests understanding of parallel circuits and how removal of one branch affects total resistance.

Explanation: This question examines the concept of internal resistance in an electrical cell. A real cell does not supply its entire emf directly to the external circuit because part of the voltage is lost inside the cell itself due to internal resistance.

The important relation involved is E=I(R+r), where E is emf, R is external resistance, r is internal resistance, and I is current. Since the same cell produces different currents with different external resistors, two equations can be formed using the same emf.

To analyze the problem, substitute the given current and resistance values into the emf relation for both situations. Because the emf remains constant, the resulting equations can be compared to eliminate the emf and solve for the internal resistance.

A practical analogy is water flowing from a tank through pipes. Some pressure is lost within the tank outlet itself before water reaches the external pipe system.

Thus, the question tests understanding of internal resistance and how current changes when the external resistance connected to a cell is varied.

Explanation: This question focuses on Ohm’s law, one of the most fundamental principles in Electricity. Ohm’s law describes the mathematical relationship between electric current, potential difference, and resistance in a conductor under constant physical conditions.

The key concept involved is that current flowing through a conductor depends directly on the applied voltage and inversely on the opposition offered by the conductor. This relationship remains valid provided temperature and other physical conditions do not change significantly.

To understand this practically, imagine water flowing through a pipe. Greater pressure difference increases water flow, while narrower pipes reduce it due to greater resistance. Electrical circuits behave similarly, with voltage acting like pressure and resistance opposing current flow.

Mathematically, Ohm’s law provides a simple proportional relation connecting these electrical quantities. It forms the basis of circuit analysis and is used extensively in designing electrical and electronic systems.

Thus, the question checks conceptual understanding of the fundamental relationship governing current flow in resistive conductors.

Explanation: This question deals with current division in a parallel circuit involving a galvanometer and a shunt resistance. A shunt is connected in parallel with the galvanometer so that most of the total current bypasses the sensitive instrument.

The important principle here is that in parallel branches, current divides inversely according to resistance. Since the shunt resistance is much smaller than the galvanometer resistance, the majority of the current naturally flows through the shunt branch.

To analyze the situation, apply the current division rule. The current through one branch depends on the resistance of the opposite branch relative to the total resistance. Using the given resistance values, the fraction of total current passing through the galvanometer can be calculated systematically.

A useful analogy is traffic choosing between a narrow crowded road and a broad highway. Most vehicles prefer the easier low-resistance route, leaving only a small fraction on the narrower path.

Thus, the question tests understanding of parallel current distribution and the practical role of shunt resistances in protecting sensitive measuring instruments.

Explanation: This question examines power dissipation in a parallel electrical circuit. In a parallel arrangement, each resistor experiences the same potential difference because both ends of every resistor are connected across the same source terminals.

The important concept involved is the relation between electrical power, voltage, and resistance. For a fixed voltage across a resistor, the power dissipated depends inversely on resistance according to the expression involving voltage and resistance. Therefore, larger resistance behaves differently compared to smaller resistance under identical voltage conditions.

To reason through the problem, compare two resistors connected in parallel where one resistance is greater than the other. Since the applied voltage across both remains equal, the resistor with smaller resistance allows more current to pass and therefore converts electrical energy into Heat at a faster rate.

A simple analogy is water flowing through two pipes connected to the same pressure source. The wider pipe allows greater flow and transfers more energy compared to the narrower pipe.

Thus, the question checks understanding of power relations in parallel circuits and how resistance affects energy dissipation under equal voltage conditions.

Explanation: This question is based on the construction of a multi-plate Capacitor. A Capacitor stores electric charge, and its capacitance depends on the effective number of plate pairs separated by insulating material.

The key concept here is that when several conducting plates are arranged alternately and connected appropriately, each adjacent pair behaves like an individual Capacitor. These individual Capacitors effectively combine in parallel, increasing the total capacitance.

To analyze the arrangement, note that if n plates are used, the number of gaps between them becomes one less than the number of plates. Since each gap contributes capacitance equal to the given value, the total capacitance becomes proportional to the number of effective gaps.

A practical analogy is stacking multiple thin sandwiches together. Each additional layer increases the total storage capacity of the stack.

Thus, the question tests understanding of how capacitance increases in multi-plate Capacitor systems and how the number of conducting plates relates to total effective capacitance.

Explanation: This question concerns Capacitors connected in series and the distribution of potential difference across them. In a series combination, each Capacitor carries the same magnitude of charge because charge cannot accumulate differently at intermediate connections.

The important concept involved is the relationship between charge, capacitance, and voltage given by Q=CV. Since the charge remains equal on both Capacitors, the voltage across each Capacitor becomes inversely proportional to its capacitance.

To analyze the situation, first determine the equivalent capacitance of the series arrangement. Then calculate the common charge stored using the total applied voltage. Finally, apply the capacitance relation individually to determine the voltage across the smaller Capacitor.

A useful analogy is two elastic balloons connected in sequence. The smaller balloon experiences greater pressure change for the same amount of stored air compared to the larger one.

Thus, the question checks understanding of charge distribution and voltage sharing in Capacitors connected in series.

Explanation: This question deals with the effective capacitance of a multi-plate Capacitor arrangement. In such systems, several conducting plates are arranged parallel to one another with equal spacing, and alternate plates are connected together.

The key idea here is that each adjacent pair of plates forms an individual capacitor. Since these Capacitors share the same potential difference, they effectively behave like Capacitors connected in parallel. Parallel combinations increase total capacitance because their individual capacities add directly.

To analyze the arrangement, count the number of spaces between the plates. If there are n plates, the number of gaps formed between adjacent plates becomes one less than the total number of plates. Each gap contributes capacitance equal to the given value.

A practical analogy is stacking multiple storage shelves together. Every extra shelf adds more storage space, increasing the overall capacity of the system.

Thus, the question tests understanding of multi-plate capacitor construction and how effective capacitance depends on the number of active plate separations.

Explanation: This question examines how different combinations of identical Capacitors produce different equivalent capacitance values. Capacitors can be connected in series, parallel, or mixed combinations, and each arrangement changes the overall capacitance of the system.

The important concept here is that series combinations reduce effective capacitance, while parallel combinations increase it. Mixed arrangements produce intermediate values depending on how the Capacitors are grouped together.

To analyze the problem, consider all unique possible configurations using three identical capacitors. One arrangement places all capacitors in series, another places them all in parallel, and additional combinations involve two in series connected parallel to the third or two in parallel connected in series with the third.

A practical analogy is combining water tanks in different ways. Depending on whether they are connected side by side or one after another, the total storage behavior changes.

Thus, the question tests understanding of capacitor combinations and how different circuit arrangements lead to distinct equivalent capacitance values.

Explanation: This question involves the relation between charge, capacitance, voltage, and charging rate in a capacitor. A capacitor stores electric charge, and the potential difference across it increases as more charge accumulates on its plates.

The key formula involved is Q=CV, where charge stored depends directly on capacitance and applied voltage. Since the capacitor is being charged steadily, the charge increases uniformly with time.

To analyze the problem, first calculate the amount of charge required to produce the specified voltage across the capacitor using the capacitance relation. Then use the given charging rate to determine how much time is needed to accumulate that quantity of charge.

A simple analogy is filling a water tank at a constant rate. The larger the tank or the higher the desired water level, the more time is required to reach that level.

Thus, the question checks understanding of capacitor charging processes and the mathematical relation connecting charge accumulation, voltage, and time.

Explanation: This question concerns the energy stored in a charged capacitor. When opposite charges accumulate on the two plates of a capacitor, electrical potential energy becomes stored in the Electric Field created between the plates.

The important concept involved is the relation between capacitance, charge, voltage, and stored energy. The energy can be expressed using different formulas depending on which quantities are known. Since charge and capacitance are provided here, the suitable energy expression can be selected accordingly.

To analyze the problem, substitute the given charge and capacitance values into the appropriate energy formula. Careful attention must be paid to unit conversion because capacitance and charge are given in microfarads and millicoulombs respectively.

A practical analogy is compressing a spring. As more compression occurs, energy becomes stored and can later be released when the spring expands.

Thus, the question tests understanding of energy storage in capacitors and the relationship between electrical charge and stored electrostatic energy.

Explanation: This question compares the equivalent capacitance obtained from series and parallel combinations of identical capacitors. The behavior of capacitors changes significantly depending on how they are connected in a circuit.

The important concept here is that capacitors connected in series produce a smaller equivalent capacitance because the effective plate separation increases. In contrast, capacitors connected in parallel increase the total capacitance because the effective plate area increases.

To analyze the situation, first recall the equivalent capacitance formulas for identical capacitors. For series combinations, the effective value becomes smaller than an individual capacitor. For parallel combinations, the capacitances add directly, producing a much larger total value.

A useful analogy is combining storage containers. Connecting them side by side increases total storage capacity, whereas arranging them sequentially restricts the effective capacity available.

Thus, the question checks conceptual understanding of how series and parallel arrangements affect total capacitance and how the two results compare quantitatively.

Explanation: This question deals with energy stored in the Electric Field between the plates of a capacitor. Energy density refers to the amount of electrical energy stored per unit volume in the region occupied by the Electric Field.

The important concepts involved are Electric Field intensity and the relation between field strength and stored energy. In a parallel plate capacitor, the Electric Field is nearly uniform and depends on the applied potential difference and plate separation.

To analyze the problem, first calculate the Electric Field using the ratio of voltage to plate distance. Then apply the standard expression for electrostatic energy density involving the permittivity of free space and the square of Electric Field intensity.

A practical analogy is compressed air stored uniformly inside a chamber. The greater the compression, the larger the energy stored in each unit volume of space.

Thus, the question tests understanding of electric fields in capacitors and how electrical energy is distributed within the space between capacitor plates.

Explanation: This question examines the energy stored in a charged capacitor. When a capacitor is connected to a voltage source, opposite charges accumulate on its plates, and electrical energy becomes stored in the Electric Field between them.

The important concept here is the relation between charge, voltage, and electrostatic energy. The stored energy depends on both the amount of charge accumulated and the potential difference across the capacitor.

To analyze the situation, apply the standard energy formula involving charge and voltage. Substitute the given values carefully while ensuring consistency of units. The result represents the total energy that would be released during discharge, usually in the form of Heat or electrical work.

A useful analogy is a stretched elastic band. The greater the stretching force and deformation, the larger the energy stored and released when allowed to return to its normal state.

Thus, the question tests understanding of electrostatic energy storage and the quantitative relationship between voltage, charge, and stored electrical energy.

Explanation: This question concerns the discharge of a charged capacitor through a resistor. When the plates of a charged capacitor are connected by a conducting path, the stored electrical energy begins to decrease as current flows through the resistor.

The important concept here is conservation of energy. Initially, the capacitor stores electrostatic energy in the Electric Field between its plates. During discharge, this energy is converted mainly into Heat in the resistor due to Joule heating.

To analyze the situation, first calculate the energy initially stored in the capacitor using the standard electrostatic energy expression involving capacitance and voltage. Since the capacitor eventually becomes completely discharged, the entire stored energy is released as Heat in the resistor. The resistance value affects the rate of discharge but not the total energy released.

A practical analogy is a compressed spring connected to a friction surface. The spring’s stored energy eventually transforms completely into Heat through frictional effects.

Thus, the question tests understanding of energy conversion during capacitor discharge and the relation between stored electrical energy and Heat produced.

Explanation: This question examines where energy is actually stored in a charged capacitor. A capacitor consists of two conducting plates separated by an insulating medium. When connected to a voltage source, equal and opposite charges accumulate on the plates.

The important concept involved is the Electric Field created between the oppositely charged plates. The work done in separating charges and building this electric field becomes stored as electrostatic potential energy.

To understand this physically, imagine stretching a rubber sheet between two points. The energy is not concentrated at the edges alone but distributed throughout the stretched region. Similarly, in a capacitor, energy is associated with the electric field existing in the space between the plates.

Modern electromagnetic theory explains that fields themselves carry energy. Therefore, the region between capacitor plates where the electric field exists becomes the actual storage location of electrostatic energy.

Thus, the question checks conceptual understanding of electric fields and the physical interpretation of energy storage in capacitor systems.

Explanation: This question is based on the relation between charge and electrostatic energy stored in a capacitor. A capacitor stores energy because work must be done to transfer charges onto its plates against the electric potential already developed.

The key concept involved is the mathematical dependence of stored energy on charge and capacitance. For a fixed capacitance, the energy expression contains the square of the charge term. This means energy does not increase linearly when charge increases.

To analyze the situation, write the energy formula in terms of charge and capacitance. Then replace the original charge with the new value and compare the resulting expression with the initial stored energy. Because the charge term is squared, doubling the charge causes a much larger increase in energy.

A practical analogy is kinetic energy in mechanics, where doubling speed increases energy much more rapidly because velocity also appears as a squared quantity.

Thus, the question tests understanding of the mathematical dependence of capacitor energy on charge and the significance of squared relationships in Physics.

Explanation: This question examines the nature of electrostatic forces and work done in moving charges within an electric field. A stationary point charge produces a radial electric field directed outward from the charge.

The important concept involved is that electrostatic forces are conservative. In a conservative field, the work done depends only on the initial and final positions, not on the actual path followed. If the starting and ending points are identical, the NET work becomes zero.

To analyze the situation, observe that moving a unit positive charge along a circular path around the stationary charge keeps the distance from the source constant throughout the motion. The electric force always acts radially, while the displacement along the circle is tangential. Since force and displacement remain perpendicular at every point, no work is performed.

A useful analogy is carrying an object around a perfectly level circular track. Although motion occurs continuously, there is no change in height and hence no NET gravitational work.

Thus, the question tests understanding of conservative electric fields and the geometrical relation between force direction and displacement.

Explanation: This question concerns the effect of dielectric materials on capacitance. A dielectric is an insulating substance inserted between the plates of a capacitor to increase its ability to store electric charge.

The important concept here is that the capacitance of a parallel plate capacitor becomes multiplied by the dielectric constant of the medium placed between the plates. The dielectric reduces the effective electric field and allows more charge to be stored for the same applied voltage.

To analyze the situation, compare the capacitance values before and after inserting the oil. The ratio of capacitance with dielectric to capacitance in air gives the dielectric constant of the material.

A practical analogy is increasing the storage capacity of a container by introducing a material that allows denser packing without overflow. Similarly, a dielectric enhances the charge storage capability of the capacitor.

Thus, the question tests understanding of dielectric materials and how they influence capacitance in electrostatic systems.

Explanation: This question examines how capacitance changes when both plate separation and dielectric medium are altered simultaneously. The capacitance of a parallel plate capacitor depends directly on dielectric constant and plate area, and inversely on plate separation.

The important relation involved is C=
d
kε
0
​

A
​

, where k is dielectric constant and d is plate separation. Increasing the distance between plates reduces capacitance, while inserting a dielectric increases it.

To analyze the problem, first express the original capacitance in terms of geometric quantities. Then modify the expression by replacing the original separation with the doubled value and introducing the dielectric constant. Comparing the new capacitance with the given multiple of the original capacitance allows the dielectric constant to be determined.

A useful analogy is widening the gap between two communicating tanks while simultaneously improving the material that helps Fluid transfer between them.

Thus, the question tests understanding of how geometrical changes and dielectric insertion jointly affect capacitor capacitance.

Explanation: This question concerns capacitance calculation when a dielectric slab partially fills the space between capacitor plates. A dielectric changes the electric field distribution and therefore modifies the effective capacitance of the system.

The important concepts involved are plate area, effective separation, and dielectric constant. When only part of the gap is filled with dielectric material, the remaining space still contains air. The system behaves as if different dielectric regions are arranged in series along the field direction.

To analyze the situation, first calculate the area of the circular plates using the given radius. Then determine the effective separation by considering the contribution of both air and dielectric regions. The dielectric slab effectively reduces the electrical thickness of the region because electric field behavior changes inside the material.

A practical analogy is replacing part of a long narrow road with a smoother surface that allows easier movement compared to the rough remaining section.

Thus, the question tests understanding of partial dielectric insertion and its influence on the effective capacitance of parallel plate capacitors.

Explanation: This question examines how inserting a dielectric medium affects the electric field inside a capacitor. A dielectric becomes polarized in the presence of an electric field, producing induced charges that oppose the original field between the plates.

The important concept involved is that the NET electric field inside the dielectric becomes smaller than the original field in air. The reduction factor depends on the dielectric constant of the material.

To analyze the situation, recall that the electric field inside a dielectric medium becomes inversely proportional to the dielectric constant. Therefore, if the dielectric constant increases, the field strength decreases accordingly.

A practical analogy is placing a shock-absorbing cushion between two moving surfaces. The cushion reduces the effective force transmitted across the gap. Similarly, the dielectric weakens the effective electric field between capacitor plates.

This reduction in electric field also contributes to the increase in capacitance because the capacitor can now store more charge for the same applied potential difference.

Thus, the question tests understanding of dielectric polarization and its effect on electric field intensity inside capacitors.

Explanation: This question deals with the behavior of a capacitor when a dielectric material is inserted while the capacitor remains connected to a constant voltage source. Under constant potential difference, the battery continues to maintain the same voltage across the plates.

The important concept here is that inserting a dielectric increases the capacitance of the condenser. Since capacitance becomes larger while voltage remains unchanged, the relation Q=CV shows that the stored charge must also change accordingly.

To analyze the situation, note that the dielectric reduces the effective electric field between the plates through polarization. This allows the capacitor to hold more charge without increasing the potential difference. The additional charge is supplied directly by the connected battery.

A useful analogy is enlarging the storage ability of a container while keeping the water pressure constant. The container naturally accommodates more stored water under the same pressure condition.

Thus, the question tests understanding of dielectric effects under constant voltage conditions and the relation between capacitance, charge, and potential difference.